Wastewater Tools: Activated Sludge and Energy Use Analysis

Transcription

1 Wastewater Tools: Activated Sludge and Energy Use Analysis Larry W. Moore, Ph.D., P.E. University of Memphis June 22, 2017

2 Objectives of Biological Treatment Oxidize dissolved and particulate biodegradable constituents into acceptable end products Capture suspended and nonsettleable colloidal solids into a biological floc or biofilm Transform or remove nutrients such as N and P Remove specific trace organic compounds Primary reference: Metcalf & Eddy 4 th Edition

3 Roles of Microbes in Wastewater Treatment Oxidation of organic matter in wastewater is accomplished biologically using a variety of microorganisms, primarily bacteria Organics + O 2 + NH 3 + PO 4 microbes New cells + CO 2 + H 2 O New cells represent the biomass produced as a result of oxidation of the organic matter

4 Energy use for a given-size WWTP may vary significantly depending on: Location Strength of wastewater Level of treatment In-plant recovery Type of treatment process Mode of operation

5 Relative distribution of energy use at a secondary wastewater treatment* plant: Relative Distribution of Energy by Process Process Post Aeration/Cl2 Mixer Lighting Heating Solids Dewatering Utility Water Effluent Filters Thickener, P.S. Secondary Clarifier RAS Activated Sludge Primary P.S., Clarifier Headworks Raw Water P.S * Sample 7.5 MGD WWTP % of Total WWTP Energy Use

6 Comments about Activated Sludge Mechanical equipment is used to provide mixing and oxygen transfer Mixed liquor flows to secondary clarifier where biomass is separated from the treated wastewater and is thickened Settled biomass is returned to aeration tank to continue biodegradation of influent organic material

7 Modeling Activated Sludge Using Bio-kinetics Dr. Moore s model uses basic design calculations based on Monod kinetics. It is a spreadsheet model using a steady-state analysis. The model is being modified using Visual Basic to create a user friendly analytical tool. The user friendly model will be available in August, 2017.

8 Activated Sludge: Basics of Design Q, S o, X o (Q-Q w ), S e, X e (Q+Q r ), S e, X v V, S e, X v Blower Q r, S e, X r Q w, S e, X r

9 Biological Reactor with Aerated Mixed Liquor (diffused aeration)

10 Key Design & Operating Parameter: θ c = MCRT = SRT = sludge age It is how long in days (on average) the biomass stays in the activated sludge system until the biomass exits the system as waste activated sludge solids or as TSS in the effluent.

11 Determining θ c Using Plant Data Q, S o, X o (Q+Q r ), S e, X v (Q-Q w ), S e, X e V, S e, X v Blower Q r, S e, X r Q w, S e, X r θ c = Q w X r + XV (Q Q w )X e Q XV w X r

12 The Activated Sludge Process 1 μ = + θ c k e θ c = mean cell residence time or sludge age μ = specific growth rate of biomass

13 Activated Sludge Design µ max = maximum specific growth rate K s = saturation constant k e = microbial decay coefficient (k e = k d ) Y k = biomass yield constant = maximum specific substrate utilization rate µ max = Yk

14 Determining S e Using Biokinetic Approach S e = θ c K s (μ (1+ k max e k θ e c ) ) 1 S e does not include CBOD 5 contributed by solids. This equation is only valid for Monod kinetics.

15 Activated Sludge Design Step 1: Determine effluent requirement CBOD 5eff = S e + f X e where X e = TSS in final effluent S e = soluble CBOD 5 f = g CBOD 5 /g TSS = 0.3 to 0.6 (Q+Q r ), S e, X v (Q-Q w ), S e, X e Q w, S e, X r

16 Activated Sludge Design Step 3: Select MLVSS concentration in aeration basin. Xv = 1500 to 3000 mg/l for complete mix 1500 to 4000 mg/l for extended air Step 4: Determine aeration basin volume V = QY X x/s v θc(s (1+ k o e θ c S ) e )

17 MLSS versus SRT 1.0 mgd Extended Aeration Act. Sludge SRT (days)

18 Activated Sludge Design Step 5: Determine the mass of volatile solids to be wasted (P XVSS ) P XVSS = A + B + C A + B = biomass production = VSW A = heterotrophic biomass B = cell debris C = nonbiodegradable VSS in influent P XVSS = QY(S o S e ) + f d (k d )YQ (S o S e ) θ c 1 + k d θ c 1 + k d θ c + QX oi

19 Activated Sludge Design Step 6: Determine the mass of total solids to be wasted (P XTSS ) P XTSS = A/ B/ C + Q(TSS o - VSS o ) where P XTSS TSS o VSS o = net waste activated sludge produced each day, mass/day = influent TSS concentration = influent VSS concentration

20 Sludge Production (TSS) vs SRT mgd Extended Aeration Act. Sludge Solids Production (lb/day) SRT (days)

21 Activated Sludge Design Step 7: Determine the oxygen requirements (CBOD and NBOD) So Se O2(lb/day) = 8.34Q 1.42(VSW) (N ox )(Q)(8.34) *Note: VSW = biomass production = A + B in previous equations 1.42(VSW) = ultimate CBOD that goes to cell growth

22 Oxygen Required (Carb+Nit) vs SRT 1.0 mgd Extended Aeration Act. Sludge O 2 (lb/day) SRT (days)

23 Activated Sludge Aeration Overall, aeration devices used for the activated sludge system represent the most significant consumers of energy within a WWTP. Most aeration systems are classified as: Diffused Dispersed Mechanical

24 The ability of any type of equipment to dissolve oxygen within a wastewater treatment system depends on: Type of aeration equipment Mixed liquor DO Basin geometry Diffuser depth Turbulence Ambient air pressure Temperature Spacing and placement of the aeration devices Wastewater characteristics Diurnal variations in wastewater flow and organic load

25 Typical Standard O 2 Transfer Rates Pump type aerators 2.3 to 3.2 lb O 2 /(HP-hr) Aspirating aerators 1.8 to 2.2 lb O 2 /(HP-hr) Horizontal rotor aerators 2.3 to 3.0 lb O 2 /(HP-hr) Nonporous diffusers 1.7 to 2.4 lb O 2 /(HP-hr) Porous diffusers 2.8 to 3.6 lb O 2 /(HP-hr)

26 Standard Conditions for Mechanical Aerators Elevation = sea level Temperature = 20 C Initial DO concentration = zero mg/l Tap water

27 Additional Assumptions for Diffused Aeration Compressor efficiency = 75% Tank depth = 15 ft Diffusers located 1.5 ft above tank bottom

28 Determine Field O 2 Transfer Rate for Mechanical Aerators OTR = OTR (βρc C) 9.2 s (T standardα ) OTR standard = oxygen transfer rate at 20 o C (lb O 2 hp -1 hour -1 ), 1 atm, tap water, and initial DO = zero mg/l C = dissolved oxygen level in basin (typically 1.5 to 2 mg/l) C s = saturated dissolved oxygen level in mg/l α = (K L a of wastewater)/(k L a of tap water); use α = 0.80 to 0.90 unless specified otherwise. β = C s wastewater/c s tap water = 0.92 for municipal wastewater ρ = factor that corrects for elevation differences

29 Approximate Field O 2 Transfer Rates Pump type aerators 1.4 to 2.0 lb O 2 /(HP-hr) Aspirating aerators 1.1 to 1.3 lb O 2 /(HP-hr) Horizontal rotor aerators 1.4 to 1.8 lb O 2 /(HP-hr) α = 0.84, β = 0.92, ρ = 1, DO = 2 mg/l, Elevation < 500 ft

30 Approximate Field O 2 Transfer Rates Nonporous diffusers 1.0 to 1.5 lb O 2 /(HP-hr) Porous diffusers 1.7 to 2.2 lb O 2 /(HP-hr) α = 0.84, β = 0.92, ρ = 1, DO = 2 mg/l Elevation < 500 ft, Compressor efficiency = 75% Tank depth = 15 ft, Diffusers located 1.5 ft above tank bottom

31 Case Study: Columbia WWTP Screens Grit Removal Preaeration Primary Clarifiers Recirculation Aeration Tanks Secondary Clarifiers Biotower Final Clarifiers RAS UV Disinfection Effluent WAS

32 Case Study: Columbia WWTP Total average daily flow rate Aeration volume in service Sec. influent BOD 5 concentration Sec. influent BOD 5 mass loading Biomass inventory (MLVSS) 4.0 mgd (half to each aer tank) 1.25 mil gal (0.625 mil gal each) 71 mg/l 2400 lb/day (total) 11,000 lb (in aeration tanks)

33 Case Study: Columbia WWTP TSS Sludge Production TSS in activated sludge effluent 1700 lb/day (intentional wastage) 330 lb/day (unintentional wastage) Oxygen Requirements for Act Sldg (actual) Oxygen required for aerobic digestion = 2.2 x VSS destroyed Total Oxygen Requirements (actual) 4000 lb/day 1200 lb/day 5200 lb/day

34 Case Study: Columbia WWTP Total Oxygen Supplied Mixing intensity in aeration tanks with 450 hp 29,000 lb/day 180 hp/mil gal

35 Columbia WWTP Estimated Energy Savings Energy savings = 150 hp x 24 hr/day x 30 day/mo x 0.75 kwh/hp-hr 81,000 kwh per month Energy cost savings = 81,000 kwh per month x $0.0866/kWh = $7,015 per month